Let \(s,p,q,q^{\prime }\)  and \(m\)  be as in Theorem REF .
Let \(n = q^{\prime }\)  and \(m = p\) . By (REF ) of Theorem REF  we have
\(\begin{split}\nu _{s}(u^{np}) \equiv \underset{r \mid t \mid n}{\sum } \mu (r) \cdot \Big ( \underset{y^{{npr}/{t}} = x^{p} \sim s}{\underset{x^{G} , \ \exists y \in G:}{\sum }} \nu _{x}(u^{n}) \Big )\pmod {p}.\end{split}\) 
Let \(k\)  be a positive integer coprime to the exponent \(e\)  of \(G\) .
Choose \(p\)  such that \(p \equiv k \pmod {e}\) . There are infinitely many such primes by Dirichlet's theorem on arithmetic progressions [1]}.
Since \(p \equiv k \pmod {e}\) , in (REF ) we have \(y^{\frac{npr}{t}} = y^{\frac{nkr}{t}}\)  and \(x^{p}=x^{k}\) . Moreover, \(u^{np}=u^{nk}\)  by the Cohn-Livingstone Theorem [2]}. This yields
\(\begin{split}\nu _{s}(u^{nk}) \equiv \underset{r \mid t \mid n}{\sum } \mu (r) \cdot \Big ( \underset{y^{{nkr}/{t}} = x^{k} \sim s}{\underset{x^{G} , \ \exists y \in G:}{\sum }} \nu _{x}(u^{n}) \Big )\pmod {p}.\end{split}\) 
The absolute value of every partial augmentation of \(G\)  is at
most \(\sqrt{|G|}\)  (really \(\nu _y(x)^2\le |y^G|\) ) by
[3]} or [4]}.
The number of summands on the right-hand side of (REF ) is at most \((2\sqrt{n})^2\cdot |G|\) . Choose \(p\)  such that \(p> (2\sqrt{n})^2\cdot |G|^{3/2}\) . Since both sides of the congruence (REF ) have absolute value less than \(p\) ,
\(\begin{split}\nu _{s}(u^{nk}) = \underset{r \mid t \mid n}{\sum } \mu (r) \cdot \Big ( \underset{y^{(knr)/t} = x^{k} \sim s}{\underset{x^{G} , \ \exists y \in G:}{\sum }} \nu _{x}(u^{n})\Big ).\end{split}\) 
If \(k\)  and \(n\)  are both congruent to 1 modulo \(|u|\) , then we get (REF ).
